Skip to content
Question

Q.The corner points of the feasible region in the graphical representation of a linear programming problem are (2,72)(2, 72), (15,20)(15, 20) and (40,15)(40, 15). If z=18x+9yz = 18x + 9y is the objective function, then :
(A) zz is maximum at (2,72)(2, 72) and minimum at (15,20)(15, 20).
(B) zz is maximum at (15,20)(15, 20) and minimum at (40,15)(40, 15).
(C) zz is maximum at (40,15)(40, 15) and minimum at (15,20)(15, 20).
(D) zz is maximum at (40,15)(40, 15) and minimum at (2,72)(2, 72).

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
Appeared in past exams:CBSE 2025· 1mCBSE 2023· 1m
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

In a linear programming problem, the maximum and minimum of the objective function occur at corner points of the feasible region. Evaluating z=18x+9yz = 18x + 9y at the given points shows the maximum is at (40,15)(40, 15) and the minimum at (15,20)(15, 20).

The Corner Point Theorem (also called the Fundamental Theorem of Linear Programming) tells us that if a linear programming problem has an optimal solution (a maximum or minimum), that solution must occur at one of the corner points (vertices) of the feasible region. This is because the objective function is linear, and the feasible region is a convex polygon — the function’s value changes linearly as you move across the region, so the extreme values will always be at the boundaries, specifically at the corners.

So, to find where zz is maximum and where it is minimum, we don’t need to graph anything or solve inequalities. We simply plug each corner point into z=18x+9yz = 18x + 9y and compare the results.

  1. Evaluate at (2,72)(2, 72)

    z=18(2)+9(72)=36+648=684z = 18(2) + 9(72) = 36 + 648 = 684

  2. Evaluate at (15,20)(15, 20)

    z=18(15)+9(20)=270+180=450z = 18(15) + 9(20) = 270 + 180 = 450

  3. Evaluate at (40,15)(40, 15)

    z=18(40)+9(15)=720+135=855z = 18(40) + 9(15) = 720 + 135 = 855

Now compare the three values:

  • 684684 at (2,72)(2, 72)
  • 450450 at (15,20)(15, 20)
  • 855855 at (40,15)(40, 15)

The largest is 855855 at (40,15)(40, 15) — that’s the maximum.

The smallest is 450450 at (15,20)(15, 20) — that’s the minimum. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.